Question

M=3/2+3/6+3/12+...+3/99*100

Answer 1

`3/2+3/6+3/12+...+3/(99*100)`

`=3(1/(1*2)+1/(2*3)+1/(3*4)+...+1/(99*100))`

`=3(1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100)`

`=3(1/1-1/100)`

`=3(100/100-1/100)`

`=3*99/100`

`=297/100`

Answer 2

Lời giải:
$M=3(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{99.100})$

$=3(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100})$

$=3(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{100-99}{99.100})$

$=3(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100})$

$=3(1-\frac{1}{100})=3.\frac{99}{100}=\frac{297}{100}$